Question

Help me I am a stone when it comes to absolute value in problems—
Explination too please

Answer 1

when dealing with an absolute value expression, let's keep in mind that the expression once we remove the bars it has a ± version, so if the expression were say hmmm |a + b| = 25, if we remove the bars we really have a duet, namely ±(a + b) = 25, same if we had |a + b| > 25, then ±(a + b) > 25.

Tis noteworthy, that unlike "equalities", inequalities vary mainly on that, when we divide or multiply or exponentialize by a negative value, the inequality sign does a horizontal flip, it changes, so let's proceed

`\boxed{31}\\\\ |b-5|\geqslant 1\implies \pm(b-5)\geqslant 1\implies \begin{cases} +(b-5)\geqslant 1\\ b\geqslant 6\\[-0.5em] \hrulefill\\ -(b-5)\geqslant 1\\ b-5\stackrel{\stackrel{notice}{\downarrow }}{\leqslant} -1\\ b \leqslant 4 \end{cases}~\hfill 6~~\leqslant ~~ b~~\leqslant ~~4`

`\boxed{32}\\\\ |-2r|\leqslant 10\implies \pm(-2r)\leqslant 10\implies \begin{cases} +(-2r)\leqslant 10\\ -2r\leqslant 10\\ r\stackrel{\stackrel{notice}{\downarrow }}{\geqslant} \cfrac{10}{-2}\\\\ r\geqslant -5\\[-0.5em] \hrulefill\\ -(-2r) \leqslant 10\\ 2r \leqslant 10\\ r\leqslant \cfrac{10}{2}\\\\ r \leqslant 5 \end{cases}~\hfill -5~~\leqslant~~ r ~~\leqslant~~5`

Check the picture below.

Answer 2

31.) you put a dot on one and go to the right

32.) you put a dot on ten and go to the left

Explanation:

the absolute value part is honestly just to throw u off, it also helps if you read it aloud. so for example in # 31. is saying the absolute value of b minus 5 is greater or equal to 1, you could also think of it as any number is greater or equal to one. hope that helps!